Poisson's and Laplace's Equations
So far we found the field from the charges. Often we face the reverse: we know the voltages on conductors but not the charge between them. Two differential equations — Poisson's and Laplace's — solve that problem directly, finding the potential from its boundaries.
- How Poisson's equation follows from Gauss's law and \(\vec{E}=-\nabla V\).
- How it reduces to Laplace's equation in charge-free regions.
- The uniqueness theorem — why any valid solution is the solution.
- A general procedure for boundary-value problems.
- Solving Laplace's equation in Cartesian, cylindrical and spherical coordinates.
- Recovering capacitance and resistance from the potential solution.
Deriving the Equations
Two results from earlier chapters combine into one powerful equation. Gauss's law in point form says \(\nabla\cdot\vec{D}=\rho_v\); with \(\vec{D}=\varepsilon\vec{E}\) and \(\vec{E}=-\nabla V\), substitute one into the other:
For a uniform medium, \(\varepsilon\) comes out of the divergence, and \(\nabla\cdot\nabla V = \nabla^2 V\) is the Laplacian of Chapter 3. This gives Poisson's equation — a single second-order equation linking the potential to the charge that produces it:
Laplace in Charge-Free Space
In most practical problems we work in the empty space between conductors, where there is no free charge: \(\rho_v=0\). Poisson's equation then loses its right-hand side and becomes Laplace's equation:
A potential satisfying \(\nabla^2 V=0\) is called harmonic (Chapter 3). Its defining property is the mean-value behaviour: the potential at any point equals the average over a surrounding sphere. Two consequences follow at once — a harmonic potential has no local maxima or minima inside a region (extremes sit on the boundaries), and the field lines neither begin nor end in the empty region.
The Uniqueness Theorem
Why can we simply guess a solution? The uniqueness theorem guarantees that any potential satisfying Laplace's (or Poisson's) equation and matching the boundary conditions is the one and only solution. There is no second, different answer hiding elsewhere.
This licenses every clever shortcut: integrate, separate variables, guess by symmetry, or use the method of images (Chapter 10). However you arrive at a potential that satisfies the equation and the boundaries, the theorem certifies it as correct.
The general procedure for a boundary-value problem is therefore fixed: (1) solve \(\nabla^2 V=0\) for the geometry, leaving integration constants; (2) apply the known potentials on the boundaries to pin those constants; (3) get \(\vec{E}=-\nabla V\); (4) if needed, find \(\rho_S=D_n\) on conductors and hence \(Q\), \(C\), or \(R\).
Cartesian Solutions
When the potential varies in only one Cartesian direction — say between two large parallel plates — Laplace's equation collapses to an ordinary second derivative, solved by two integrations:
Applying \(V=V_0\) at \(x=0\) and \(V=0\) at \(x=d\) fixes the constants, giving \(V = V_0(1-x/d)\). The field \(\vec{E}=-dV/dx\,\hat{a}_x = (V_0/d)\hat{a}_x\) is uniform — recovering the parallel-plate result of Chapter 8 from the boundaries alone, with no mention of charge.
Cylindrical Solutions
For a coaxial geometry where \(V\) depends only on \(\rho\), the Laplacian keeps just its radial part. Setting it to zero and integrating twice gives a logarithmic potential:
With \(V=V_0\) on the inner conductor (\(\rho=a\)) and \(V=0\) on the outer (\(\rho=b\)), the potential and field become
From \(\rho_S=\varepsilon E\) on the inner conductor we recover the coaxial capacitance \(C=2\pi\varepsilon L/\ln(b/a)\) of Chapter 8 — again, starting from voltages rather than charge.
Spherical Solutions
For concentric spheres where \(V\) depends only on \(r\), the radial Laplacian gives a \(1/r\) potential:
Matching \(V_0\) at \(r=a\) and \(0\) at \(r=b\) and differentiating recovers the spherical capacitor \(C=4\pi\varepsilon ab/(b-a)\). The table below collects the three one-dimensional solution forms — knowing these three shapes (linear, logarithmic, reciprocal) handles the bulk of undergraduate boundary-value problems:
| Geometry | Reduced equation | General solution |
|---|---|---|
| Cartesian (in \(x\)) | \(d^2V/dx^2=0\) | \(V=Ax+B\) |
| Cylindrical (in \(\rho\)) | \(\tfrac{1}{\rho}\tfrac{d}{d\rho}(\rho\,V')=0\) | \(V=A\ln\rho+B\) |
| Spherical (in \(r\)) | \(\tfrac{1}{r^2}\tfrac{d}{dr}(r^2 V')=0\) | \(V=A/r+B\) |
Worked Examples
Problem. Does \(V = x^2 - y^2 + 2z\) satisfy Laplace's equation?
Solution. Sum the second derivatives:
Problem. Plates at \(x=0\) (\(100\ \text{V}\)) and \(x=2\ \text{mm}\) (\(0\ \text{V}\)). Find \(V(x)\) and \(\vec{E}\).
Solution. Use \(V=Ax+B\) with the boundary values:
Problem. A region \(0\le x\le d\) has uniform \(\rho_v\), with \(V(0)=0\), \(V(d)=0\). Find \(V(x)\).
Solution. Integrate \(d^2V/dx^2=-\rho_v/\varepsilon\) twice and apply boundaries:
Problem. A coaxial line has \(a=2\ \text{mm}\), \(b=8\ \text{mm}\), inner at \(60\ \text{V}\). Find \(V\) at \(\rho=4\ \text{mm}\).
Solution. Use \(V=V_0\ln(b/\rho)/\ln(b/a)\):
Problem. Show the coaxial solution reproduces \(C=2\pi\varepsilon L/\ln(b/a)\).
Solution. Get \(\rho_S=\varepsilon E\) at \(\rho=a\), integrate for \(Q\), divide by \(V_0\):
Problem. Concentric spheres \(a=1\ \text{cm}\) (\(50\ \text{V}\)), \(b=3\ \text{cm}\) (\(0\ \text{V}\)). Find \(V(r)\).
Solution. Use \(V=A/r+B\) and apply both boundary values:
Chapter Summary
\(\nabla^2 V=-\rho_v/\varepsilon\); follows from Gauss's law plus \(\vec{E}=-\nabla V\).
\(\nabla^2 V=0\) in charge-free regions; solution is harmonic, extremes on the boundary.
Any solution fitting the equation and boundaries is the only solution — guess freely.
Solve, apply boundaries, get \(\vec{E}=-\nabla V\), then \(\rho_S\), \(Q\), \(C\) or \(R\).
Linear \(Ax+B\); logarithmic \(A\ln\rho+B\); reciprocal \(A/r+B\).
Recovers plate, coaxial and spherical capacitance from voltages, not charge.
Problems
For each item, pick the coordinate that matches the symmetry, write the reduced equation, then apply the boundary values. Difficulty rises down the list.
- Check whether \(V = 3x + 4y\) and \(V = xy\) satisfy Laplace's equation.
- Plates at \(x=0\) (\(0\ \text{V}\)) and \(x=5\ \text{mm}\) (\(200\ \text{V}\)). Find \(V(x)\) and \(\vec{E}\).
- Solve Poisson's equation for uniform \(\rho_v\) between grounded plates at \(x=0\) and \(x=d\).
- A coaxial line (\(a=1\ \text{mm}\), \(b=5\ \text{mm}\)) has inner at \(100\ \text{V}\). Find \(V\) and \(\vec{E}\) at \(\rho=2\ \text{mm}\).
- Concentric spheres (\(a=2\ \text{cm}\), \(b=6\ \text{cm}\)) with inner at \(80\ \text{V}\). Find \(V(r)\).
- From the spherical solution, derive \(C=4\pi\varepsilon ab/(b-a)\).
- Two coaxial cylinders carry inner \(V_0\), outer grounded. Find the surface charge density on each.
- A region between \(x=0\) and \(x=a\) has \(\rho_v=\rho_0 x/a\). Solve Poisson's equation with grounded ends.
- Show that a harmonic potential can have no maximum or minimum inside a charge-free region.
- Two long coaxial conductors are filled with a leaky dielectric (\(\sigma\)). Find the leakage resistance per unit length and relate it to \(C\).
- A potential satisfies \(\nabla^2 V=0\) with \(V=0\) on all boundaries of a closed region. Prove \(V=0\) everywhere inside.
- Explain why the uniqueness theorem makes the method of images (next chapter) legitimate.